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A horizontal beam safely supports a load \( p \) that varies jointly as the product of the width \( w \) of the beam and the square of the depth \( d \), and inversely as the length \( l \).

A beam with a width of 6 inches, a depth of 24 inches, and a length of 6 inches can safely support 6,336 pounds.

Determine the safe load in pounds of a beam made from the same material if the beam is 18 inches long.

a) 1,408 lb
b) 2,112 lb
c) 2,816 lb
d) 3,520 lb

Answer :

Using the given joint variation equation, the safe load for a beam 18 inches in length made from the same material is determined to be 352 pounds, hence none of the provided options are correct.

The load p varies jointly with the product of the width w and the square of the depth d, and inversely with the length l. This relationship can be represented by the equation

p = k * (w * d^2) / l,

where k is a proportionality constant. We are given that a beam with a width of 6 inches, depth of 24 inches, and length of 6 inches can safely support 6,336 pounds.

Hence, we can write

6,336 = k * (6 * 24^2) / 6.

Simplifying, we find

k = 6,336 / (6 * 24^2).

To find the load that a beam of 18 inches length can support, we can set l = 18 and solve for p while the width and depth remain constant.

Substituting the known values in the equation gives us

p = (6,336 / (6 * 24^2)) * (6 * 24^2) / 18,

hence p = 6,336 / 18 = 352 pounds.

Therefore, the safe load that the beam can support if it is 18 inches long is 352 pounds.

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